Q19 is not 15, I managed to find a counter example.Let m=2 and solve for a1, you get a1^3-2a1^2+a1-1=0. Plug it into a cubic equation solver, you get 1.754877666246692 as one of the roots. Solve for 16a1-a3, you get somethung like 23.998 which is clearly more than 15, so 15 can't be the answer.

Q19 should be 24. First notice that the terms from a2 onwards are positive. If 01 and the rest of the terms are positive, so no solution. If a1 is negative, you have 16a1-a_{m+1} is negative, so we can ignore this case. Plug small values of m into WolframAlpha, we have 16a1-a_{m+1}=15, 23.9984, - 6.5..., - -1765...,.... This probably means 24 is the answer as 16a1-a_{m+1} decreases as m increases for m>2.

Q19 I also got 24. Manipulation of the recurrence relation can get 1/(a_{n})=1/(a_{n}-1)+1/(a_{n+1}-1). Plug into 1/a1+...+1/am =1, we can get 1/(a_{1}-1)-1/(a_{m+1}-1)=1 by method of difference. Then we can express a_{m+1} in terms of a1, which is 1/(2-a_{1}). Substitute it into 16a1-am+1 and differentiate. maximum value is reached when 2-a1=1/4. Sub in get 24. I think this process is correct but a1=1.75 is clearly not correct. This is absurd... but 15 is clearly wrong given the counter-example